Product of roots of $ax^2 + (a+3)x + a-3 = 0$ when these are positive integers

Question

There is only one real value of $'a'$ for which the quadratic equation $$ax^2 + (a+3)x + a-3 = 0$$ has two positive integral solutions.The product of these two solutions is :

Since the solutions are positive, therefore the product of roots and sum of root will be positive. This will give us two inequalities in $a$. Substitute the values of $a$ in the quadratic but I'm not getting my answer correct.

The two inequalities will be $$\frac{a-3}{a} > 0$$ and the other one will be $$\frac {a+3}{a} < 0$$

From the first one we get $a>3$ and from the second one we get $a< -3$.

I don't know how to proceed after this.

Kindly help.

Answer 1

Thanks to @rtmd, I fixed the solution. This only holds when $a=-\frac{3}{7}$. Here is the proof.

$\frac{a-3}{a}, \frac{a+3}{a} \in \mathbb{Z}$ implies $\frac{6}{a} \in \mathbb{Z}$. We can write $a=\frac{6}{n}$ for some integer $n$. \begin{align*} \frac{a-3}{a}= 1-\frac{n}{2} \in \mathbb{Z}\\ \frac{a+3}{a}= 1+\frac{n}{2} \in \mathbb{Z} \end{align*} Therefore $n$ is even. Assume $p,q$ are the positive integer solutions of the given quadratic equation. \begin{align*} pq= 1-\frac{n}{2} \in \mathbb{Z}\\ p+q= -1-\frac{n}{2} \in \mathbb{Z} \end{align*} This gives us \begin{align*} pq-p-q&=2\\ (p-1)(q-1)&=3 \end{align*} Therefore WLOG $p=4,q=2$. This only holds when $n=-14$ and $a=-\frac{3}{7}$

Answer 2

We need $$\frac{a+3}{a}\in\Bbb{Z}\ , \frac{a-3}{a}\in\Bbb{Z}$$ or $$1+\frac{3}{a}\in\Bbb{Z}\ , \ 1-\frac{3}{a}\in\Bbb{Z}$$ thus $\displaystyle \frac{3}{a}\in\Bbb{Z}$, means that $\displaystyle a=\frac{3}{m}$ where $m\in\Bbb{Z}$.

Now we can write the equation as $$\frac{3}{m}x^2+\left(\frac{3}{m}+3\right)x+\frac{3}{m}-3=0\implies x^2+(m+1)x+1-m=0$$ Using quadratic formula we have $$x_{1,2}=\frac{-m-1\pm\sqrt{m^2+6m-3}}{2}=\frac{-m-1\pm\sqrt{(m+3)^2-12}}{2}$$We need the expression under the square root to be an integer. There are only two squares with difference $12$ (those are $4,16$), hence we want $(m+3)^2=16\implies m=1\text{ or }m=-7$, thus $$a=3 \text{ or } a=-\frac{3}{7}$$ and as $a<0$ we have

$$a=-\frac{3}{7}$$